An early model for an atom considered it to have a positively charged nucleus of charge zeze surrounded by a uniform density of negative charge up to a radius rr. The atom as a whole is neutral. For this model, what is the electric field at a distance rr from the nucleus, where r>rr>r?
Answers
Answer:
Final Answer :
When r <=R, E(r) = 0
When r >R, E(r) = Ze/4π£° (1/r^2 - r/R^3)
STEPS AND Understanding :
1) We know, that size of Nucleus in negligible as compared to Nucleus.
So, At any distance r, we have to understand there exists nucleus with charge( +Ze. )
2) As atom is Neutral, the total Negative Charge in a sphere of Radius R is (-Ze. )
3) We will find Negative charge density to calculate charge at any distance r.
4) We will. apply Gauss law to find Electric Field at less than equal to R distance and more than that.
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Answer:
Explanation:
Charge on nucleus = + Ze=+Ze
Total negative charge = -Ze=−Ze
(\because(∵ atom is electrically neutral)
Negative charge density, \rho = \dfrac {charge}{volume} = \dfrac {-Ze}{\dfrac {4}{3}\pi R^{3}}ρ=
volume
charge
=
3
4
πR
3
−Ze
i.e., \rho = -\dfrac {3}{4} \dfrac {Ze}{\pi R^{3}} ...... (i)ρ=−
4
3
πR
3
Ze
......(i)
Consider a Gaussian surface with radius rr.
By Gauss's theorem,
E(r) \times 4\pi r^{2} = \dfrac {q}{\epsilon_{0}} ..... (ii)E(r)×4πr
2
=
ϵ
0
q
.....(ii)
Charge enclosed by Gaussian surface,
q = Ze + \dfrac {4\pi r^{3}}{3} \rho = Ze - Ze \dfrac {r^{3}}{R^{3}}q=Ze+
3
4πr
3
ρ=Ze−Ze
R
3
r
3
(Using (i))
From (ii),
E (r) = \dfrac {q}{4\pi \epsilon_{0}r^{2}} = \dfrac {Ze - Ze\dfrac {r^{3}}{R^{3}}}{4\pi \epsilon_{0}r^{2}} = \dfrac {Ze}{4\pi \epsilon_{0}}\left [\dfrac {1}{r^{2}} - \dfrac {r}{R^{3}}\right ]E(r)=
4πϵ
0
r
2
q
=
4πϵ
0
r
2
Ze−Ze
R
3
r
3
=
4πϵ
0
Ze
[
r
2
1
−
R
3
r
].